Given a smooth affine scheme $X$, let's consider $\varinjlim\limits_{U\subset X}\tilde{K}_0(U)$, where $\tilde{K}_0(U)$ is the reduced zero-th $K$-group of the Zariski open subscheme $U$. Note that since algebraic vector bundles are Zariski locally trivial this implies that this colimit is zero. Specially this implies that $\varinjlim\limits_{U\subset X}CH^i(U)\otimes \mathbb{Q}=0$. Here $CH^i(U)$ is the Chow group of codimension $i$ cycles. The colimit $\varinjlim\limits_{U\subset X}CH^i(U)$ is also equal to zero. Since for any cycles restricting to its complement, will send the cycles to zero by localization.
Is it possible to get an idea what the $\varinjlim\limits_{U\subset X}C^d(U)$ looks like when the base field is $\mathbb{C}$? where $C^d(U)$ is the $Mor(U,C_d(Y))$. Here $C_d(Y)$ is the Chow variety of codimension $d$ cycles of a fixed variety $Y$ and the colimit is taken as the colimit of topological spaces.
For $i>0$: If $Z$ is a codimension $i$ for $i>0$ irreducible subvariety in $U$, certainly under the restriction map $$CH^i(U)\to CH^i(U-Z)$$ $$Z\mapsto 0$$ So in the colimit $\varinjlim\limits_{U\subset X}CH^i(U)$, $Z\mapsto 0$.
For $i=0$: if $X$ is irreducible, then $\varinjlim\limits_{U\subset X}CH^0(U)\simeq \mathbb{Z}.$